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RD Sharma Class 12 Exercise 17.3 Maxima And Minima Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 17.3 Maxima And Minima Solutions Maths - Download PDF Free Online

Updated on Jan 21, 2022 02:35 PM IST

By regularly practicing RD Sharma Class 12th Exercise 17.3, understudies will be severe with the thoughts given in the Class 12 RD Sharma textbook. Be that as it may, it is nearly impossible to be skilled at maths who thought proper practice. The Rd Sharma class 12th exercise 17.3 will help students to learn and practice at home so that they have the chance to improve their skills to score well in boards and JEE mains.

This Story also Contains
  1. RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise
  2. Maxima and Minima Excercise: 17.3
  3. Maxima and Minima excercise 17.3 question 1 (iii)
  4. Maxima and Minima Excercise 17.3 question 1 (v)
  5. Maxima and Minima exercise 17.3 question 1 (vi)
  6. Maxima and Minima excercise 17.3 question 1 (ix)Answer:
  7. Maxima and Minima exercise 17.3 question 2 (i)
  8. Maxima and Minima exercise 17.3 question 3
  9. Maxima and Minima excercise 17.3 question 4
  10. Maxima and Minima exercise 17.3 question 8
  11. Benefits of picking RD Sharma Arithmetic Solutions from Career360 include:
  12. RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise

Maxima and Minima Excercise: 17.3

Maxima and Minima excercise 17.3 question 1 (i)

Answer:

Point of local maxima is at x=1 and it’s local maximum value is 68. Also point of local minima is at x=5,-6 and it’s local minimum values are -316 and -1647.

Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If f"(c1)<0 then c1 is point of local minima.
If f"(c2)<0 then c2 is point of local maxima .
where c1 &c2 are critical points.
Put c1 and c2 in f(x) to get minimum value & maximum value
Given:
f(x)=x462x2+120x+9
Explanation:
It is given that
f(x)=x462x2+120x+9f(x)=4x3622x+120
To find critical values ,
f(x)=04x3124x+120=04(x331x+30)=0
On solving,
critical values of x are 5,1,-6.
Now,
f(x)=12x21
At x=5,
f(5)=12(5)2124=176>0
So, x=5 is point of local minima.
At x=1,
f(1)=12(1)2124=112<0
So, x=1 is point of local maxima.
At x=-6,
f(6)=12(6)2124=308>0

So, x=-6 is a point of local minima.
The local max. value at x=1 is
f(1)=(1)462(1)2+120(1)+9=68
And local min. value at x=5 & x=-6 are,
f(5)=(5)462(5)2+120(5)+9=316f(6)=(6)462(6)2+120(6)+9=1647
Hence, point of local maxima is 1 & its maximum value is 68.
Also, point of local minima are 5&-6 and its minimum values are -316 &-1647 respectively.

Maxima and Minima excercise 17.3 question 1 (ii)
Answer:

point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If f(c1)>0 then c1 is point of local minima.
If f(c2)<0then c2 is point of local maxima .
where c1 & c2 are critical points.
Put c1 and c2 in f(x) to get minimum value & maximum value.
Given:
f(x)=x36x2+9x+15
Explanation:
We have,

f(x)=x36x2+9x+15f(x)=3x26.2x+9=3x212x+9f(x)=3(x24x+3)&f(x)=3(2x4)=6x12
To find maxima and minima.
f(x)=03(x24x+3)=0x24x+3=0X=3 or x=1 At x=3,f(3)=6(3)12=6>0
At x = 3,
f’’(3) = 6(3)-12
=6>0
So, x = 3 is point of local minima
At x = 1,
f’’(1) = 6(1)-12
=-6<0
So, x = 1 is point of local maxima
Now, local maximum value at x= 1 is
f(1)=(1)36(1)2+9(1)+15f(1)=19

And local min. value at x=3 is
f(3)=(3)36(3)2+9(3)+15f(3)=15
Thus, point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.

Maxima and Minima excercise 17.3 question 1 (iii)

Answer:
Point of local maxima value is -2 and it’s local maximum value is 0. Also point of local minima is 0and it’s local minimum value is -4
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If f(c1)>0 then c1 is point of local minima.
If f(c2)<0then c2 is point of local maxima .
where c1 &c2 are critical points.
Put c1 and c2 in f(x) to get minimum value & maximum value.
Given:
f(x)=(x1)(x+2)2
Explanation:
We have,
f(x)=(x1)(x+2)2f(x)=(x+2)2(1)+(x1)2(x+2)usingddxu.v=udvdx+vdudx=(x+2)(x+2+2x2)=(x+2)(3x)f(x)=3x(1)+(x+2)3usingddxu.v=udvdx+vdudx=6x+6
For max and min, f’(x)=0,
(x+2)(3x)=0x=0&x=2
At x=0,
f(0)=6(0)+6=6>0
So, x= 0 is point of local minima
f(0)=6(0)+6=6>0
At x=-2,
f(0)=6(2)+6=6<0
So, x = -2 is point of local maxima
So, local max. value at x=-2 is
f(-2)=(-2-1)(-2+2)^{2}=0
And local min. value at x= 0 is
f(0)=(01)(0+2)2=4
Thus, point of local maxima is -2 & its max. value is 0 & point of local minima is 0 & its value is -4.

Maxima and Minima exercise 17.3 question 1 (iv)

Answer:

Point of local maxima is 2 and it’s local maximum value is 12
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If f(c1)>0 then c1 is point of local minima.
If f"(c2)<0then c2 is point of local maxima .
where c1 & c2 are critical points.
Put c1 and c2 in f(x) to get minimum value & maximum value.
Given:
f(x)=2x2x2,x>0
Explanation:
We have,
f(x)=2x2x2,x>0f(x)=2x2+4x3f(x)=4x212x4

For maxima and minima, f’(x)=0
2x2+4x3=02(x2)x3=0x=2
At x=2,
f(2)=4(2)312(2)2=481216=1234=14<0
So, x=2 is point of local maxima.
So, local max. value at x= 2 is
f(2)=222(2)2=124=12
Hence, point of local maxima is 2 & its max. value is 12

Maxima and Minima Excercise 17.3 question 1 (v)

Answer:
Point of local minima value is x=-1 and it’s local minimum value is 1e.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If f"(c1)>0 then c1 is point of local minima.
If f"(c2)<0 then c2 is point of local maxima .
where c1 & c2 are critical points.
Put c1 and c2 in f(x) to get minimum value & maximum value.
Given:f(x)=xex
Explanation:
We have,
f(x)=xexf(x)=ex(x+1) using ddxu.v=udvdx+vdudxf(x)=ex(x+1)f(x)=ex(x+1)+exusingddxu.v=udvdx+vdudxf(x)=ex(x+2)
Now, for maxima & minima, f’(x)=0
ex(x+1)=0x+1=0x=1
at x=-1,
f(1)=e1(1+2)=1e
X = -1 is point of local minima.
So, local min. value at x= 1 is
f(1)=1e1=1/e
Thus, point of local minima is -1 & its min value is 1e

Maxima and Minima exercise 17.3 question 1 (vi)

Answer:
Point of local minima value is 2 and it’s local minimum value is 2.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If f(c1)>0 then c1 is point of local minima.
If f(c2)>0 then c2 is point of local maxima .
where c1 &c2 are critical points.
Put c1 and c2 in f(x) to get minimum value & maximum value.
Given:
x2+2x,x>0
Explanation:
We have,
f(x)=x2+2x,x>0f(x)=122x2 f(x)=4x3
For local maxima or minima, we have f(x)=0
122x2=0x2=4x=2 or x=2
since, x>0
x=2
Now, at x=2 ,
f(2)=22+22=2
Thus, its min. value at 2 is 2.

Maxima and Minima Excercise 17.3 question 1 (vii)
Answer:

Point of local minima value is 74and it’s local minimum value is 3(4)43
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If f(c1)>0 then c1 is point of local minima.
If f(c2)>0 then c2 is point of local maxima .
where c1 &c2 are critical points.
Put c1 and c2 in f(x) to get minimum value & maximum value.
Given:
f(x)=(x+1)(x+2)13,x>=2
Explanation:
we have ,
f(x)=(x+1)(x+2)f(x)=(x+2)13+13(x+1)(x+2)(23) using ddxu.v=udvdx+vdudxf(x)=23(x+2)2/329(x+1)(x+2)(53) using ddxu.v=udvdx+vdudx
For maxima & minima, f’(x)=0
(x+2)13+13(x+1)(x+2)23=013(x+1)=(x+2)13(x+2)2313(x+1)=(x+2)x+1=3x6x=74
At x=74
f(74)=23(74+2)2329(74+1)(74+2)(53)=23(14)23+118(14)53>0
So , x=74 is a point of local minima
Now at x=74
f(74)=(74+1)(74+2)13=(34)(14)13=3443
Thus, point of local minima is 74 & its value is 3443

Maxima and Minima exercise 17.3 question 1 (viii)

Answer:
Point of local maxima value is 4 and it’s local maximum value is 16. Also point of local minima is -4 and it’s local minimum value is -16
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If f(c1)>0 then c1 is point of local minima.
If f(c2)>0 then c2 is point of local maxima .
where c1 &c2 are critical points.
Put c1 and c2 in f(x) to get minimum value & maximum value.
Given:
f(x)=x32x2,5x5
Explanation:
We have,
f(x)=x32x2,5x5f(x)=32x2x232x2usingddxu.v=udvdx+vdudxf(x)=x32x2(2x32x2+x332x2(32x2))usingddx(uv)=vdudx+udvdxv2
Now for local minima and local maxima
f(x)=032x2x232x2=032x2=x2x2=16x=±4
At x=4,
f(4)=43242[(8(3242)+43)(3242)(3242]=119264=3<0
So, x=4 is a point of local maxima.
At x=4,
f(4)=432(4)2=43216=16
At x=-4,
f(4)=4(4)32(4)2[8(3242)(4)3(3242)3242]=0+2=3>0
So, x=-4 is the point of local minimum.
At x=-4
f(x)=(4)32(4)2=43216=16
Thus, point of local maxima is at x=4 is and its max. value is 16 and & point of local minima is at x=-4 is and its minimum value is -16.

Maxima and Minima excercise 17.3 question 1 (ix)
Answer:

Point of local maxima value is a3 and it’s local maximum value is 4a327. Also point of local minima is a and it’s local minimum value is 0.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If f(c1)>0 then c1 is point of local minima.
If f(c2)<0 then c2 is point of local maxima .
where c1 & c2 are critical points.
Put c1 and c2 in f(x) to get minimum value & maximum value.
Given:
f(x)=x32ax2+a2x,a>0
Explanation:
We have,
f(x)=x32ax2+a2xf(x)=3x24ax+a2f(x)=6x4a
For maxima and minima f’(x) =0
3x24ax+a2=0x=(4a)±(4a)24×3×a22×3=4a±16a212a26=4a±2a6x=a or x=a3
At x= a,
f(a)=6(a)4a=2a>0
So, x=a is point of local minima
Now at x=a,
f(a)=a32a(a)2+a2(a)=a32a3+a3=0
Also, at x=a3
f(a3)=(a3)22a(a3)+a2(a3)=4a327
Thus, point of local maxima is a3 & its max. value is 4a327& point of local minima is a & its value is 0.

Maxima and Minima excercise 17.3 question 1 (x)

Answer:
Point of local minima value is a and it’s local minimum value is 2a. Also point of local maxima is -a and it’s local maximum value is -2a
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If f(c1)>0 then c1 is point of local minima.
If f(c2)<0 then c2 is point of local maxima .
where c1 &c2 are critical points.
Put c1 and c2 in f(x) to get minimum value & maximum value.
Given:
f(x)=x+a2x,a>0
Explanation:
We Have,
f(x)=x+a2xf(x)=1a2x2f(x)=2a2x3
For maxima and minima ,f’(x)=0
1a2x2=0x=±a

Maxima and minima exercise 17.3 question 1 (xi)

Answer:

x=1 is a point of local maxima and its maximum value is 1 and x=-1 is a point of local minima and its minimum value is -1.
Hint:
Putting x=1,-1in f''(x)
Given:
f(x)=x2x22x2
Explanation:
Differentiating f(x) with respect to x
f(x)=ddx(x2x2)
 Using, ddx[f(x)g(x)]=f(x)ddx{g(x)}+g(x)ddx{f(x)}f(x)=xddx2x2+2x2ddx(x)(1)
 Using, df(x)ndx=nf(x)n1ddxf(x) and ddx[f(x)±g(x)]=ddx(f(x))±ddx(g(x))ddx2x2=ddx(2x2)1/2=12(2x2)1/21ddx(2x2)=12(2x2)1/2(2x)=x2x2
From(1),f(x)=x(x2x2)+2x2×1=x2+2x22x2=22x22x2
 Put, f(x)=022x22x2=022x2=02x2=2x2=1x=±1 if x2 and x2
These x=1 and x=-1 we the possible point of local maxima and minima
Differentiating f’(x) with respect to x
f(x)=ddx(22x22x2)=2x2ddx(22x2)(22x2)ddx2x2(2x2)
Using the formula,
ddx(uv)=vddxudaxvv2=2x2×(4x)(22x2)12(2x2)1/2(2x)2x22 (Applying chain rule) =4x2x2+x(22x2)(2x2)3/2=4x(2x2)+2x2x3(2x2)3/2=8x+4x3+2x2x3(2x2)3/2
f(x) at x=1f(1)=6(1)+2(1)3(2(1)2)3/2=6+2(1)3/2=4<0

So x=1 is local maxima and its maximum value f(1)=1

f(x) at x=1f(1)=6(1)+(2)(1)3(2(1)2)3/2=62(1)3=4>0

So , x = 1 is a point of local maxima and its minimum value f(-1)=-1

Maxima and Minima Excercise 17.3 question 1 (xii)

Answer:
x=34 is a point of local maxima and its maximum value is 5/4.
Hint:
Using chain rule of Differentiation
Given:
f(x)=x+1xx1
Explanation:
Differentiating f(x) with respect to x
f(x)=ddx(x+1x)=ddx(x)+ddx(1x)1/2=1+12(1x)121(1) [ Using Chain Rule Diffrentiation ]
=1121x=21x121x Put f(x)=021x121x=021x1=0 if x11x=12 [Squaring Both Sides]
 i. e1x=14x=34
Thus , x=34 is the point of local maxima and minima
Differentiating f’ with respect to x
f(x)=ddx(1121x)=ddx(1)ddx(121x)=012ddx(1x)1/2=12×12(1x)3/2(1) (Using chain rule) =14(1x)32 Put f(x) at x=34f(34)=14(134)3/2=14×122=22<0

So x=34is a point of local maxima
The local maximum value at f(x)
f(12)=34+134=34+12=3+24=54

Maxima and Minima exercise 17.3 question 2 (i)

Answer:

x=2 is the local minima and the local minimum value of f(x) at x=2 is 0.
x=43 is the local maxima and the local maximum value of f(x)at x=43 is x=427.

Hint:
Using chain rule
Given:
f(x)=(x1)(x2)2
Explanation:
Differentiating f with respect to x
f(x)=ddx[(x1)(x2)2)=(x1)ddx(x2)2+(x2)2ddx(x1)=2(x1)(x2)+(x2)2 [Using chain rule] =2(x22xx+2)+(x24x+4)=2x24x2x+4+x24x+4=3x210x+8
 Put f(x)=03x210x+8=0x=b±b24ac2a=10±(10)24382(3)=+10±100966=10±26
x=+10+26=2 And x=+1026=43
Differentiating f(x) with respect to x
f(x)=ddx(3x210x+i)=6x10
When put x = 2 in f’’(x)
f(2)=6(2)10=2>0
So x=2 is the local Minima
The Local Minimum value of f(x) at x =2 is 0
Put x=43 in f(x)
f(2)=6(2)10=2>0
The local maximum value of f(x)at x=43 is x=427.

Maxima and Minima excercise 17.3 question 2 (iii)

Answer:
x=1 is point of inflexion
x=-1 is the point of local minimum &x=15 is the point of local maximum
Hint:
Using chain rule of derivative
Given:
f(x)=(x1)3(x+1)2
Explanation:
Differentiating f with respect to x
f(x)=ddx[(x1)3(x+1)2]=[(x1)3ddx(x+1)2+(x+1)2ddx(x1)3]=[(x1)32(x+1)+3(x+1)2(x1)2]=(x1)2(x+1)[2x2+3x+3]=(x1)2(x+1)(5x+1)=(x22x+1)(5x2+x+5x+1)=(x22x+1)(5x2+6x+1)=(5x4+6x3+x210x312x22x+5x2+6x+1)=(5x44x36x2+4x+1)
 Put f(x)=01(x1)2(x+1)(5x+1)=0(x1)2(x+1)(5x+1)=6(x1)2=0x1=0x=1x+1=0,5x+1=0x=1,x=15
Thus x=1 and x=1 and x=15 are the possible points of local minima and maxima
Differentiating f’(x) with respect to xf(x)=ddx[(5x44x36x2+4x+1)]=(20x312x212x+4)=20x3+12x2+12x4 when x=1f(1)=20(1)3+12(1)2+12(1)4=20+12+124=0

Thus this test is fail as x=1 is point of inflexion
when x=1
f(1)=20(1)3+12(1)2+12(1)4=20+12124=16>0

So , x = -1 is a point of local minimum

When x=15

f(15)=20(15)3+12(15)2+12(15)4=336/125<0
So, x=-1/5 is the point of local maximum

Maxima and Minima exercise 17.3 question 3

Answer:
a=23
b=16
Hint:
Put? dydx=0 at x=1,2
Given:
y=alogx+bx2+x
Explanation:
Differentiating y with respect to x
dydx=ax+2 bx+1=ax+2bx+1 At x=1(dydx)=a1+2 b(1)+1=a+2 b+1(1)
 At x=2dydx=a2+2(b)2+1=a2+4b+1(2)
Since x=1 and x=2 are the extreme values,f’(x)=0 at x=1 & 2
Then equation (1) and (2) become
a+2b+1=0(3)a2+4b+1=0a+8b+2=0(4) [Multiplying both sides by 2] 
Solving , (3) and (4) , we get
6b1=0b=16
Putting B=1/6in(3),weget
a13+120a=23

Maxima and Minima excercise 17.3 question 4

Answer:
We need to prove that x = e is local maxima
Given:
The given function logxx
Explanation:
 Let y=logxx
Then
dydx=xddx(logx)log(x)ddx(x)x2,ddx(uv)=vdudx+udvdxv2=x×1xlogxx2=1logxx2
Put,
dydx=01logxx2=01logx=0,x0logx=1x=e1=e
Differentiating dydx with respect to x
d2ydx2=x2ddx(1logx)(1logx)ddx(x2)x4.ddx(uv)=vdudx+udvdxv2=x2(1x)2x(1logx)x4=x2x+2xlogxx4=3x2xlogxx4
At x=e,
d2ydx2=3e+2elogee4=3e+2e(1)e4loge=1=ee4=1e3<0
So x = e is a point of local maxima

Maxima and Minima exercise 17.3 question 5

Answer:
x = 0 is local minima and its minimum value is 2 and x =1 is local maxima and its maximum value is -6
Given:
f(x)=4x+2+x
Explanation:
Differentiating f with respect to x
f(x)=ddx(4(x+2)1+x)=4(x+2)2+1=4(x+2)2+1=4+x2+4x+4(x+2)2=x2+4x(x+2)2=x(x+4)(x+2)2
 Put f(x)=0x(x+4)(x+2)2=0x(x+4)=0 i.e x=0, or x=4
Thus x = 0 and x = -4 are the points of local maxima and minima
Differentiating f’ with respect to x
f(x)=ddx(x2+4x(x+2)2)=(x+2)2ddx(x2+4x)(x2+4x)ddx(x+2)2(x+2)4ddx(uv)=vdudx+udvdxv2=(x+2)2(2x+4)(x2+4x)2(x+2)(x+2)4=2(x+2)[(x+2)2(x2+4x)](x+2)4=2(x+2)(x2+4x+4x24x)(x+2)4=8(x+2)(x+2)4=8(x+2)3
Put x = 0 in f ‘’ (x)
f(0)=8(0+2)3=88=1>0
S,x=0is local minima. Hence,its minimum value will be f(0)=2
And x= -4 in f ‘’(x)
f(4)=8(4+2)3=88=1<0
So x=-4 is local maxima. Hence, its maximum value will be f(-4)=-6

Maxima and Minima exercise 17.3 question 6

Answer:
x=π4 is a point of local minima and its local minimum value will be f(π4)=1π2
x=3π4 is a point local maxima and its local maximum value will be f(3π4)=13π2
Hint:
 Put fx=0
Given:
f(x)=tanx2x
Explanation:
Differentiating f(x) with respect to x
f(x)=sec2x2 Put f(x)=0sec2x2=0sec2x=2secx=±2=secx=2x=π4 or, secx=2 or, secx=ππ4=3π4
Thus x=π4 or x=3π4 is possible points of local maxima and minima
Differentiating f’(x) with respect to x
fu(x)=ddx[sec2x2]=2secx(secxtanx)=2sec2xtanx Put x=π4 in f(x)f=(π4)=2sec2(π4)tan(π4)=2(2)(1)=4>0
So ,
x=π4 is a point of Local Minima and its local maximum value will be f(π4)=1π2
 And Put, x=3π4 in f(x)f(3π4)=2sec2(3π4)tan(3π4)=2(2)2(1)=4<0
So x=3π4 is a point local maxima and its local maximum value will be f(3π4)=13π2

Maxima and Minima exercise 17.3 question 7
Answer:
a=3b=9&cR
Hint:
Put,
f(1)=0f(3)=0
Given:
f(x)=x3+ax2+bx+C
Explanation:
Differentiating f(x) with respect to x
f(x)=3x2+2ax+b+0=3x2+2ax+b
 Calculate f(x) at x=1 and x=3f(1)=3(1)2+2a(1)=b=32a+bf(3)=3(3)2+2a(3)+b=27+6a+b
Put and f(1)=0 and f(3)=0 as they are the extremum values hence f(x)=0
32a+b=072a+b=3(1)276a+b=206a+b=27(2)(1)(2),wc get 2a6a=3+278a=2ya=3 Put a=3 in (1)2(3)2b=3 b=9

Maxima and Minima exercise 17.3 question 8

Answer:
Hence proved.
Hint: f(x)=0
Put,
Given:
f(x)=sinx+3cosx
Explanation:
Differentiating fx with respect to x
f(x)=cosx3sinx Put f(x)=0cosx3sinx=03sinx=cosxtanx=13
x=π6
Differentiating f’(x) with respect to x ,
f(x)=sinx3cosx Putt x=π6 at f(x)f(π6)=sinπ63cosπ6=123×32=1232=42=2<0
So, x=π6 is a point local maxima.

Rd Sharma Class 12th exercise 17.3 is an essential part of the RD Sharma solutions for maths. The Class 12 RD Sharma chapter 17 exercise 17.3 solution will contain an important chapter of the NCERT maths book that needs to be studied well. RD Sharma Class 12 Solutions Maxima and Minima Ex 17.3 is going to be based on the chapter Maxima and minima.
Rd Sharma class 12-chapter 17 exercises 17.3 answers that are basic and simple. It will help students learn all formulae well and become skilled at the subject in general. RD Sharma class 12 solutions ex 17.3 has around 21 inquiries.

  • Solutions identified with the Greatest and least upsides of a function in its space
  • Definition and which means of greatest
  • Definition and significance of least
  • Nearby maxima
  • Questions identified with Nearby minima
  • Discover Solutions of First subordinate test for neighborhood maxima and minima
  • Higher-request subordinate test

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